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Solution to Exercise 4.74.

If X is $LogNormal(\mu = 1.9, \sigma = .9)$ , then

$\begin{displaymath} Y \; = \; \log X \; \sim \; N(1.9,(.9)^2).\end{displaymath}$

(a) Using the formulae on p. 181,

$\begin{displaymath} E[X] \; = \; \exp( 1.9 + (.9)^2/2 ) 10.02 ,\end{displaymath}$

and

$\begin{displaymath} V[X] \; = \; \exp(2*1.9 + (.9)^2 )*(\exp((.9)^2) -1) \; = \; 125.4 ,\end{displaymath}$

$\begin{displaymath} \sigma_X \; = \; 11.20 .\end{displaymath}$

(b)

$\begin{displaymath} P[ X \le 10 ] \; = \; P[ Y \le \log 10 ] \; = \; \Phi \le... ...{\log(10)-1.9}{.9} \right) \; = \; \Phi (.45) \; = \; .6736 .\end{displaymath}$

$\begin{displaymath} P[ 5 < X \le 10 ] \; = \; P[X \le 10 ] - P[X \le 5]\end{displaymath}$

$\begin{displaymath} \; = \; .6736 - \Phi \left( \frac{\log(5)-1.9}{.9} \right) ... ... \; .6736 - \Phi(-.32) \; = \; .6736 - .3745 \; = \; .2991 .\end{displaymath}$

Dennis Cox
2/26/2001